Question

A lamina is made by removing a small disc of diameter $2 R$ from a bigger disc of uniform mass density and radius $2 R$, as shown in the figure. The moment of inertia of this lamina about axes passing though $O$ and $P$ is $I _{0}$ and $I_{P}$ respectively. Both these axes are perpendicular to the plane of the lamina. The ratio $I_{P} / I_{O}$ to the nearest integer is

Answer 1

$I_{p}=\left[\frac{4 m R^{2}}{2}+m\left(4 R^{2}\right)\right]-\left[\frac{m}{4} \frac{R^{2}}{2}+\frac{m}{4} 5 R^{2}\right] $
$I_{p}=m R^{2}\left[(2+4)-\left(\frac{1}{8}+\frac{5}{4}\right)\right] $
$I_{p}=m R^{2}\left(6-\frac{11}{8}\right)=\frac{37}{8} m R^{2}\ldots $(1)
$I_{0}=\left(\frac{4 m R^{2}}{2}\right)-\left(\frac{m}{4} \frac{R^{2}}{2}+\frac{m}{4} R^{2}\right)$
$I_{0}=m R^{2}\left[2-\left(\frac{1}{8}+\frac{1}{4}\right)\right]=m R^{2}\left[2-\frac{3}{8}\right]=m R^{2}\left(\frac{13}{8}\right) \ldots$ (2)
So $\frac{I_{p}}{I_{0}}=\frac{37 / 8}{13 / 8} 3$ (Nearest Integer)